\( \frac{1}{10} \) of a pole is coloured red, \( \frac{1}{20} \) is white, \( \frac{1}{30} \) is blue, \( \frac{1}{40} \) is black and the rest is green. If the length of the green portion of the pole is 4.75 m, then the length of the pole is:
To find the total length of the pole, we first need to determine the fraction of the pole that is green. This can be done by subtracting the sum of the fractions of the other coloured parts from the total length (which is 1). Once we have the fraction representing the green portion, we can use the given length of the green portion to find the total length of the pole.
To add these fractions, find the Least Common Multiple (LCM) of the denominators (10, 20, 30, 40). The LCM of these numbers is 120.
\[ \text{Sum of fractions} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30} + \frac{1}{40} \]\[ = \frac{12}{120} + \frac{6}{120} + \frac{4}{120} + \frac{3}{120} \]\[ = \frac{12 + 6 + 4 + 3}{120} = \frac{25}{120} \]The total length of the pole is represented by 1. So, the green portion is the total length minus the sum of the other coloured portions.
\[ \text{Fraction of green portion} = 1 - \frac{5}{24} \]\[ = \frac{24}{24} - \frac{5}{24} = \frac{24 - 5}{24} = \frac{19}{24} \]We are given that the length of the green portion is 4.75 m.
Let \( L \) be the total length of the pole.
\[ \frac{19}{24} \times L = 4.75 \text{ m} \]\[ L = \frac{4.75 \times 24}{19} \]\[ L = \frac{114}{19} \]\[ L = 6 \text{ m} \]B) 6 m
The question asks for the year when the name 'Pakistan' was coined. This is a factual question related to the history of the Indian subcontinent and the movement for a separate Muslim state.
Correct Option: A) 1933 is the correct year when the name 'Pakistan' was coined by Choudhry Rahmat Ali.
The question asks to identify the second person to land on the Moon. This requires knowledge of the Apollo 11 mission and the sequence of astronauts who stepped onto the lunar surface.
D) Buzz Aldrin was the second person to land on the Moon, following Neil Armstrong during the Apollo 11 mission.
We need to find the ratio of the area of the square formed by the centers of the four circles to the area of the rhombus formed by their touching points. We will first determine the side length of the square and the diagonals of the rhombus in terms of the radius \(r\).
Let the radius of each circle be \(r\). When four circles are placed such that their centers form a square, the distance between the centers of any two adjacent circles is \(r + r = 2r\). This distance is the side length of the square formed by the centers.
Let the side length of the square be \(s_1\).
\[ s_1 = 2r \]The area of this square is:
\[ A_{\text{square}} = s_1^2 = (2r)^2 = 4r^2 \]The four circles touch each other. The touching points form a rhombus. Consider the square formed by the centers. The touching points lie on the midpoints of the sides of this square.
Let's visualize the centers as A, B, C, D forming a square. The touching point between circle A and circle B is the midpoint of the line segment AB. Similarly for other pairs.
The vertices of the rhombus are the midpoints of the sides of the square formed by the centers. This means the rhombus is formed by connecting the midpoints of the sides of the square ABCD.
The diagonals of this rhombus are equal to the side length of the square formed by the centers. Let the diagonals of the rhombus be \(d_1\) and \(d_2\).
The length of each diagonal of the rhombus will be equal to the side length of the square formed by the centers, which is \(2r\).
\[ d_1 = 2r \] \[ d_2 = 2r \]The area of a rhombus is given by the formula:
\[ A_{\text{rhombus}} = \frac{1}{2} d_1 d_2 \]Substituting the values of \(d_1\) and \(d_2\):
\[ A_{\text{rhombus}} = \frac{1}{2} (2r)(2r) = \frac{1}{2} (4r^2) = 2r^2 \]Now, we need to find the ratio of the area of the square to the area of the rhombus:
\[ \text{Ratio} = \frac{A_{\text{square}}}{A_{\text{rhombus}}} = \frac{4r^2}{2r^2} \] \[ \text{Ratio} = \frac{4}{2} = \frac{2}{1} \]So, the ratio is \(2 : 1\).
Correct Option: D) 2 : 1
The question asks for the year when the Asiatic Cheetah was officially declared extinct in India. This is a factual recall question related to wildlife conservation history in India.
B) 1952 — The Asiatic Cheetah was officially declared extinct in India in 1952.
The question asks for the establishment year of Tata Iron and Steel Company Limited (TISCO). This is a factual question related to Indian history and industrial development. Recalling the historical timeline of major industrial ventures in India is key to answering this question.
Correct Option: D) 1907. Tata Iron and Steel Company Limited (TISCO), now known as Tata Steel, was founded by Jamsetji Tata and established in 1907 at Jamshedpur, Jharkhand (then part of Bengal Presidency).
The problem is a blood relations puzzle. We need to deduce the relationships between the 9 family members based on the given clues and then identify the wife of the doctor. We will build a family tree step-by-step.
Correct Option: C) Q is the wife of the doctor (N).
If x =\(\frac{\sqrt{3}+1}{\sqrt{3}-1}\) and y = \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\), then the value of \( x^{2} + y^{2} \) is:
To find the value of \(x^2 + y^2\), we first need to simplify the expressions for \(x\) and \(y\) by rationalizing their denominators. After simplifying \(x\) and \(y\), we can then calculate \(x^2\) and \(y^2\) and sum them up. Alternatively, we can use the identity \(x^2 + y^2 = (x+y)^2 - 2xy\).
Given \(x = \frac{\sqrt{3}+1}{\sqrt{3}-1}\). Multiply the numerator and denominator by the conjugate of the denominator, which is \((\sqrt{3}+1)\).
\[ x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \]\[ x = \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2 - (1)^2} \]\[ x = \frac{(\sqrt{3})^2 + 2(\sqrt{3})(1) + (1)^2}{3 - 1} \]\[ x = \frac{3 + 2\sqrt{3} + 1}{2} \]\[ x = \frac{4 + 2\sqrt{3}}{2} \]\[ x = 2 + \sqrt{3} \]Given \(y = \frac{\sqrt{3}-1}{\sqrt{3}+1}\). Multiply the numerator and denominator by the conjugate of the denominator, which is \((\sqrt{3}-1)\).
\[ y = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \]\[ y = \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - (1)^2} \]\[ y = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + (1)^2}{3 - 1} \]\[ y = \frac{3 - 2\sqrt{3} + 1}{2} \]\[ y = \frac{4 - 2\sqrt{3}}{2} \]\[ y = 2 - \sqrt{3} \]Now that we have \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), we can calculate their squares.
\[ x^2 = (2 + \sqrt{3})^2 \]\[ x^2 = (2)^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 \]\[ x^2 = 4 + 4\sqrt{3} + 3 \]\[ x^2 = 7 + 4\sqrt{3} \]\[ y^2 = (2 - \sqrt{3})^2 \]\[ y^2 = (2)^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2 \]\[ y^2 = 4 - 4\sqrt{3} + 3 \]\[ y^2 = 7 - 4\sqrt{3} \]Add the values of \(x^2\) and \(y^2\).
\[ x^2 + y^2 = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) \]\[ x^2 + y^2 = 7 + 4\sqrt{3} + 7 - 4\sqrt{3} \]\[ x^2 + y^2 = 7 + 7 \]\[ x^2 + y^2 = 14 \]A) 14 is the correct answer, as calculated by simplifying \(x\) and \(y\) and then finding the sum of their squares.
If sin A + sin B = x and cos A + cos B = y, then cos(A - B) = ?
\(\frac{x^2+y^2+2}{2}\)
\(\frac{x^2-y^2+2}{2}\)
\(\frac{x^2-y^2-2}{2}\)
\(\frac{x^2+y^2-2}{2}\)
We are given two equations involving sums of sines and cosines of angles A and B. We need to find an expression for \(\cos(A-B)\) in terms of \(x\) and \(y\). The key idea is to square both given equations and add them, then use trigonometric identities to simplify the expression.
Given the first equation:
\[ \sin A + \sin B = x \quad \text{(1)} \]
Square both sides of equation (1):
\[ (\sin A + \sin B)^2 = x^2 \]
\[ \sin^2 A + \sin^2 B + 2 \sin A \sin B = x^2 \quad \text{(3)} \]
Given the second equation:
\[ \cos A + \cos B = y \quad \text{(2)} \]
Square both sides of equation (2):
\[ (\cos A + \cos B)^2 = y^2 \]
\[ \cos^2 A + \cos^2 B + 2 \cos A \cos B = y^2 \quad \text{(4)} \]
Add equation (3) and equation (4):
\[ (\sin^2 A + \sin^2 B + 2 \sin A \sin B) + (\cos^2 A + \cos^2 B + 2 \cos A \cos B) = x^2 + y^2 \]
Rearrange the terms:
\[ (\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2 (\sin A \sin B + \cos A \cos B) = x^2 + y^2 \]
Apply the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ 1 + 1 + 2 (\cos A \cos B + \sin A \sin B) = x^2 + y^2 \]
\[ 2 + 2 \cos(A - B) = x^2 + y^2 \]
Here, we used the cosine difference formula: \(\cos(A - B) = \cos A \cos B + \sin A \sin B\).
Solve for \(\cos(A - B)\):
\[ 2 \cos(A - B) = x^2 + y^2 - 2 \]
\[ \cos(A - B) = \frac{x^2 + y^2 - 2}{2} \]
D) \(\frac{x^2+y^2-2}{2}\) is the correct expression derived from the given equations.
The question asks to identify the Indian state to which the Paika Dance belongs. This requires knowledge of traditional Indian folk dances and their regional origins.
The Paika dance is a form of martial folk dance performed by the Paikas (warriors) of Odisha. It is characterized by the use of shields, swords, and other weapons, depicting the valor and martial prowess of the ancient warriors.
C) Odisha: The Paika Dance is a well-known traditional martial folk dance originating from the state of Odisha. It is performed by the Paikas, who were a class of warriors, and involves acrobatic movements and the use of traditional weapons like swords and shields.